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A satellite is orbiting Earth at a distance h = R_{E} above the surface. What is the centripetal acceleration of the satellite compared to that on the surface?
A. a = g/4
B. a = g/2
C. a = g ^ 2
D. a = 2g E, a = 4g
Please walk through it to I’m so lost



Answer :

Answer:

Can you write the height again please as i just see R_{E}

In any case though you know the formula for centrepetal acceleration

which is the acceleration due to gravity is

g = GM/r^2

where r is the distance between the object and Earth

so if the sattelite is R+h distance above than earth

so g = GM/(r+h)^2

hence according to the R value g will change by square factor

as if the disntace change is 2 times g will change by 4 times

g and R are indirectly proportional

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